Q.47. The de Broglie wave length for an electron accelerated through the potential difference volt

Q: What is the correct answer for this exam Physics Q47?

The correct answer is D. The de Broglie wavelength for an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$. So $\lambda \propto \frac{1}{\sqrt{V}}$. Given $\lambda_1 = \frac{h}{\sqrt{2meV_1}}$. When the potential difference is $V_2$, the wavelength $\lambda_2$ is increased by 50%, so $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$. So $\frac

Q: Q.47. The de Broglie wave length for an electron accelerated through the potential difference volt is . When the potential difference is changed to volt, the associated de Broglie wavelength is…

The correct answer is D. The de Broglie wavelength for an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$. So $\lambda \propto \frac{1}{\sqrt{V}}$. Given $\lambda_1 = \frac{h}{\sqrt{2meV_1}}$. When the potential difference is $V_2$, the wavelength $\lambda_2$ is increased by 50%, so $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$. So $\frac

Question

Q.47. The de Broglie wave length for an electron accelerated through the potential difference  volt is . When the potential difference is changed to  volt, the associated de Broglie wavelength is increased by 50%. If , then the value of  is _______.

FirstInTest · Accepted Answer

The de Broglie wavelength for an electron accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$. So $\lambda \propto \frac{1}{\sqrt{V}}$. Given $\lambda_1 = \frac{h}{\sqrt{2meV_1}}$. When the potential difference is $V_2$, the wavelength $\lambda_2$ is increased by 50%, so $\lambda_2 = \lambda_1 + 0.5\lambda_1 = 1.5\lambda_1 = \frac{3}{2}\lambda_1$. So $\frac{\lambda_2}{\lambda_1} = \frac{3}{2}$. We have $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{V_1}{V_2}}$. So $\frac{3}{2} = \sqrt{\frac{V_1}{V_2}}$. Squaring both sides, $\frac{9}{4} = \frac{V_1}{V_2}$. Given $\frac{V_1}{V_2} = \frac{9}{\alpha}$. So $\frac{9}{4} = \frac{9}{\alpha}$. This implies $\alpha = 4$.

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